A boiler is made of a copper plate `2.4mm` thick with an inside coating of a `0.2mm` thick layer of tin The surface area exposed to gases at `700^(@)C` is `400cm^(2)` The maximum amount of steam that could be generated per hour at atmospheric pressure is
`({:(K_(cu)=0.9cal//cm-s-^(0)& k_("tin")=0.15cal//cm-s-^(0)C andL_(steam)=540cal//g:})` .

Options

A. `5000Kg`
B. `1000Kg`
C. `4000Kg`
D. `200Kg`

Explanation

To solve the problem of determining the maximum amount of steam that could be generated per hour in a boiler made of copper and tin, we will follow these steps:

### Step 1: Identify the given data
- Thickness of copper plate, \( L_{cu} = 2.4 \, \text{mm} = 0.24 \, \text{cm} \)
- Thickness of tin layer, \( L_{tin} = 0.2 \, \text{mm} = 0.02 \, \text{cm} \)
- Conductivity of copper, \( k_{cu} = 0.9 \, \text{cal/cm-s-°C} \)
- Conductivity of tin, \( k_{tin} = 0.15 \, \text{cal/cm-s-°C} \)
- Surface area exposed to gases, \( A = 400 \, \text{cm}^2 \)
- Temperature outside the boiler, \( T_{outside} = 700 \, \text{°C} \)
- Temperature inside the boiler, \( T_{inside} = 100 \, \text{°C} \)
- Latent heat of steam, \( L_{steam} = 540 \, \text{cal/g} \)

### Step 2: Calculate the total thermal resistance
The total resistance \( R_s \) for heat transfer through the two layers (copper and tin) can be calculated using the formula:

\[ R_s = R_{cu} + R_{tin} = \frac{L_{cu}}{k_{cu} \cdot A} + \frac{L_{tin}}{k_{tin} \cdot A} \]

Substituting the values:

\[ R_s = \frac{0.24 \, \text{cm}}{0.9 \, \text{cal/cm-s-°C} \cdot 400 \, \text{cm}^2} + \frac{0.02 \, \text{cm}}{0.15 \, \text{cal/cm-s-°C} \cdot 400 \, \text{cm}^2} \]

Calculating each term:

1. For copper:
\[ R_{cu} = \frac{0.24}{0.9 \cdot 400} = \frac{0.24}{360} = 0.0006667 \, \text{s-°C/cal} \]

2. For tin:
\[ R_{tin} = \frac{0.02}{0.15 \cdot 400} = \frac{0.02}{60} = 0.0003333 \, \text{s-°C/cal} \]

Now, adding them together:
\[ R_s = 0.0006667 + 0.0003333 = 0.001 \, \text{s-°C/cal} \]

### Step 3: Calculate the heat current (I)
Using the formula for heat transfer:
\[ I = \frac{V}{R_s} \]
Where \( V \) is the temperature difference:
\[ V = T_{outside} - T_{inside} = 700 - 100 = 600 \, \text{°C} \]

Substituting the values:
\[ I = \frac{600 \, \text{°C}}{0.001 \, \text{s-°C/cal}} = 600000 \, \text{cal/s} \]

### Step 4: Calculate the mass of steam generated per second
Using the latent heat to find the mass of steam generated:
\[ \text{Mass of steam per second} = \frac{I}{L_{steam}} = \frac{600000 \, \text{cal/s}}{540 \, \text{cal/g}} \approx 1111.11 \, \text{g/s} \]

### Step 5: Convert to mass of steam generated per hour
To find the total mass of steam generated in one hour:
\[ \text{Mass per hour} = 1111.11 \, \text{g/s} \times 3600 \, \text{s} = 4000000 \, \text{g} = 4000 \, \text{kg} \]

### Final Answer
The maximum amount of steam that could be generated per hour at atmospheric pressure is **4000 kg**.

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